February 21, 2013

It surprised me to learn that the West Indies recently won their first match in Australia against Australia since early 1997. Over 32 matches (11 tests, 19 one day internationals, 2 Twenty 20 internationals) prior to the win they had two no results in ODIs and one draw in the tests. In the first of the no-result games they were in a winning position, reducing Australia to 5/43 chasing 264 before rain washed out the game.

A list of the matches can be found via ESPN CricInfo’s Statsguru here. The best batting and bowling performances by the Windies during that time were as follows.

Brian Lara scored two test centuries (including a 226 in Adelaide) and one ODI century.

Chris Gayle scored two test centuries including 165* in the drawn game and the fifth fastest ever test century (in balls faced).

Dwayne Bravo also scored two test centuries and had the best test bowling figures of 6/84.

Like this:

February 20, 2013

In this post we look at the sum two sinusoids of different amplitude and phase but where one has twice the frequency of the other. How should we choose the phase so that the sum has minimal peak?

Mathematically, given a constant we wish to find

The sum can take a variety of forms, two of which are shown in red below ( and ).

We claim that a value of for which the peak value of is maximised is as illustrated below (again ).

We see here the two sinusoids reach their minimum at , resulting in a minimum value of for the sum. However the maximum value is only for the case . Hence we make the observation that finding the lowest peak is different from finding the minimum deviation from zero.

I initially approached this problem using the method of Lagrange multipliers. Consider the problem of finding the maximum value of where and are fixed. We write

where are variables and are constants. Hence we wish to maximise subject to . By the method of Lagrange multipliers, at the maximum the gradient vector of the objective function is parallel to the gradient of the constraint. In other words, . Hence there exists a constant so that

Solving these two equations for and gives and . Then the condition gives us the quartic equation

By multiplying (1) by and (2) by and adding, we obtain

where is one of the solutions of (3).

Unfortunately only for a few special cases does the solution in have a “nice” form. So this approach to finding the best was not appealing.

After conjecturing that the best is , one alternative approach is to proceed as follows. We firstly note that for ,

Hence the maximum value is or depending on how large is.

Now we wish to show that for other values of there exists for which this maximum value is exceeded. Firstly note that the more interesting case is since when , we can achieve a minmax value of by aligning the peak of with the trough of (e.g. for ), obtaining a sum value of at . This is at the local maximum since here,

and

.

From now on we consider .

Choose such that (i.e. is kept constant). Then ,

and

. Then

Write this as , and note that . We wish to show if there exists close to such that . We use the fact that near , while . In particular for all sufficiently close (but not equal) to ,

where . Furthermore choose so that (i.e. ) and . We then find

It follows that

,

or , where can be chosen to be positive for some . Hence we have as desired and

Hence we would have an for which the claimed maximum value of has been exceeded. We conclude that we must have and (e.g. when ), leading to the minmax value of .

Recapping, we have shown that

It is interesting to see that the behaviour of the minmax changes from being quadratic to linear in as exceeds . For the examples plotted above, , leading to a minmax value of as we found in the third graph.

If we wish to find where , we simply write this as where and use the above result to obtain the following: